2  Complex-valued functions

Definition 2.1 A complex function \(f\colon D\to\C\) is a map with domain of definition \(D\subset\C\) and codomain the complex plane. Thus, \(f\) assigns to each \(z=x+iy\in D\) in the domain a complex number\[f(z)=u(z)+iv(z).\]We call \(u\colon D\to\R\) the real part and \(v\colon D\to\R\) the imaginary part of the complex function \(f.\)

Figure 2.1: Schematic picture of a complex function

We can visualize complex functions as in Figure 2.1. In practice, most complex functions are defined by a formula.

Example 2.1  

Let \(a,b\in\C\) be fixed complex numbers. Then \(f(z)=az+b\) is a complex function with domain \(D=\C,\) written \(f\colon\C\to\C.\) For instance, \[f(z)=iz, f(z)=2z, f(z)=z+3.\]

Example 2.2  

A polynomial is a complex function \(f\colon \C\to\C\) of the form \[f(z)=a_nz^n+a_{n-1}z^{n-1}+\ldots+a_1z+a_0, \tag{2.1}\] with complex coefficients \(a_n,\ldots, a_0\in\C\) and with domain \(D=\C.\)

More generally, a rational function is a complex function for which there are polynomials \(f,g\) with \(g\neq0\) so that\[h(z)=\frac{f(z)}{g(z)}. \tag{2.2}\]

The domain of \(h\) is \(D=\{z\in\C\mid g(z)\neq0\}.\) However, if the polynomials \(f,g\) have a common factor, we can cancel that factor in the fraction Equation 2.2 and regard \(h\) as a complex function on a larger domain.

For example, \(h(z)=\frac{z+1}{z^2-1}\) has domain \(\C\setminus\{\pm1\},\) but we can rewrite the fraction as \(h(z)=\frac{1}{z-1},\) which makes sense on the extended domain \(\C\setminus\{+1\}.\)

We have already met several complex functions in the previous section.

Example 2.3  

The exponential function \(\exp\colon\C\to\C\) is defined by \(\exp(z)=e^z=e^x(\cos(y)+i\sin(y))\) and has domain \(D=\C.\)

Example 2.4  

The sine function \(\sin\colon\C\to\C\) and the cosine function \(\cos\colon\C\to\C\) are the complex functions defined by \[\cos(z)=\frac{e^{iz}+e^{-iz}}{2}, \qquad \sin(z)=\frac{e^{iz}-e^{-iz}}{2i}. \tag{2.3}\] (See the corresponding exercise in Section 1.1)

Definition 2.2 We define the following subsets of the complex plane: \[\begin{align*} \C^\t&=\{z\in\C\mid z\neq0\} &&\text{\textbf{punctured plane}}\\ \C^-&=\C\setminus\{x\in\R\mid x\leqslant 0\} &&\text{\textbf{slit plane}}\\ S&=\{z=x+iy\mid y\in(-\pi,\pi)\} &&\text{\textbf{principal strip}}\\ \H&=\{z=x+iy\mid y>0\}&&\text{\textbf{upper half-plane}} \end{align*}\]

Figure 2.2: The principal strip \(S\) and the slit plane \(\C^-\)

Example 2.5  

The principal branch of the complex logarithm is the complex function

\[\log\colon \C^-\longra S\subset\C\]defined by\[\log(w)=\log(r)+i\th\iff w=re^{i\th}, r>0, \th\in(-\pi,\pi). \tag{2.4}\]

Here \(\log(r)\) denotes the (real) logarithm from MA1005 Calculus. In other words, the real part of \(\log(w)\) is the logarithm of the modulus \(r=|w|\) and the imaginary part of \(\log(w)\) is the argument function \(\arg(w)\) from Definition 1.3.

To evaluate Equation 2.4, write \(w\) in polar coordinates, ensuring that \(\th\in(-\pi,\pi).\) For example, \(i=e^{i\pi/2}\) and so \(\log(i)=\log(1)+i\frac{\pi}{2}=i\frac{\pi}{2}.\)

Proposition 2.1  

1. The exponential function is surjective onto \(\C^\t.\)

2. We have \[\exp(\log(w))=w\ (\forall w\in\C^-) \tag{2.5}\] \[\log(\exp(z))=z\ \; (\forall z\in S). \tag{2.6}\]

   Hence the restriction \(\exp|_S\) of the exponential to the principal strip is a bijection \(\exp|_S\colon S\to\C^-\) onto the slit plane, with inverse \(\log(w).\)

Proof.

1. Recall from MA1005 Calculus that \(e^x\neq0\) for all \(x\in\R.\) As \(|\exp(x+iy)|=e^x\neq0,\) the image of \(\exp\) is contained in \(\C^\t.\) For proving that \(\exp\) is onto \(\C^\t,\) recall that every non-zero number can be written in polar form \(w=re^{i\th},\) \(\th\in(-\pi,\pi].\) Then \(\exp(z)=w\) for \(z=\log(r)+i\th.\)

2. Equations Equation 2.5, Equation 2.6 are straightforward to verify using Equation 2.4.

Since the graph \(\Ga(f)=\{(z,w)\in D\t\C\mid f(z)=w\}\) of a complex function is a subset of four-dimensional space, we cannot visualize complex functions as easily as real functions. We will now discuss some alternatives.

Image grid

A useful way to picture a complex function is to sketch its values on a grid \(G.\) The image grid \(f(G)\) is a distorted version of the original grid which can be navigated easily using the grid lines. For example, to determine \(f(1+2i),\) take one step in \(x\)-direction and two steps in \(y\)-direction on the distorted grid. Formally, let \(G=\{z=x+iy\in\C\mid x\in\Z\text{ or }y\in\Z\}\) be the unit grid and define the image grid as (see Figure 2.3)

\[f(G)=\{w=u+iv\in\C\mid\exists z\in G: f(z)=w\}.\]

Figure 2.3: The unit grid \(G\) and the image grid \(f(G)\) for \(f(z)=\frac{1+i}{\sqrt{2}}z\)

In practice, the image grid can often be described by finding a familiar equation that all of its members \(u+iv\) satisfy. When this is not possible, a computer will help sketching an approximate image grid.

Example 2.6  

Consider \(f(z)=z^2.\) Then \[u=x^2-y^2,\qquad v=2xy. \tag{2.7}\] To determine \(f(G),\) first fix \(x=\pm1,\pm2,\pm3,\ldots\) to be a non-zero integer. Using Equation 2.7 we find \(u=\frac{-1}{4x^2}v^2+x^2.\) This is a downward parabola in the \((u,v)\)-plane rotated by 90 degrees with vertex at \((u,v)=(x^2,0)=(1,0),(4,0),(9,0),\ldots.\) Similarly, if we fix \(y\) to be a non-zero integer, then \(u=\frac{1}{4y^2}v^2-y^2\) is an upward parabola in the \((u,v)\)-plane rotated by 90 degrees. For \(x=0,\) we get \((u,v)=(-y^2,0)\) which parameterizes the negative \(u\)-axis. Similarly, for \(y=0\) we get the positive \(u\)-axis. All this is summarized in Figure 2.4.

Figure 2.4: Unit grid and image grid of \(f(z)=z^2\)

Example 2.7  

Consider \(f(z)=1/z.\) Then \[u=\frac{x}{x^2+y^2},\qquad v=\frac{-y}{x^2+y^2}. \tag{2.8}\] Fixing \(x=\pm1,\pm2,\ldots,\) we have \(\bigl(u-\frac{1}{2x}\bigr)^2+v^2=\frac{1}{4x^2}\) (verify by substituting Equation 2.8 into this equation). This is a circle of radius \(\frac{1}{2x}\) and with center \((u,v)=(\frac{1}{2x},0).\) For \(x=0,\) equations Equation 2.8 become \((u,v)=(0,-1/y)\) which parameterizes the \(v\)-axis. Similarly, for \(0\neq y\in\Z\) we find \(u^2+\bigl(v+\frac{1}{2y}\bigr)^2=\frac{1}{4y^2}\) and for \(y=0\) we obtain a parametrization of the \(u\)-axis. This is summarized in Figure 2.5 (and on the title page).

Figure 2.5: Unit grid and image grid of \(f(z)=1/z.\) Notice that \(z=0\) gets sent to a ‘point at infinity’ that is imagined to surround the complex plane.

The previous example can be generalized.

Example 2.8  

Rational functions Equation 2.2 with \(f(z)=az+b\) and \(g(z)=cz+d\) affine linear, where \(a,b,c,d\in\C,\) are Möbius transformations. Thus

\[f(z)=\frac{az+b}{cz+d}\]

with domain \(D=\C\setminus\{-d/c\}\) if \(c\neq0\) and domain \(D=\C\) if \(c=0.\) To exclude constant functions, we also assume that \(ad-bc\neq0.\)

Domain colouring

We represent each unit complex number \(e^{i\th}\) by a color on the color wheel. The modulus \(r\) of an arbitrary complex number \(re^{i\th}\) will be represented by the lightness of the color. This assigns a unique color to each complex number, see Figure 2.6. Pure white is never used and would correspond to infinity. Pure black corresponds to the origin.

This is less useful for calculations by hand but generates artistic images using a computer.

Figure 2.6: Representing complex numbers \(z\) by color

This can be used for visualizing complex functions. Draw each point \(z\) in the domain of \(w=f(z)\) using the color for \(w.\)

Figure 2.7: Domain colouring of \(f(z)=\frac{(z-2)(z-1)^2}{z^2+1}.\) Notice that near \(z=\pm i,\) where \(f\) is undefined, the function tends to infinity (pure white). The zeros of \(f\) at \(z=1,2\) can also be seen.

3-dimensional graphs

Another approach is to plot the 3-dimensional graph of any of the following real-valued functions \(D\to \R\) \[u, v, |f|=\sqrt{u^2+v^2}.\]

Again, the missing information can be color-coded (see Figure 2.8).

Figure 2.8: The graph of \(|f(z)|\) colored by \(\arg(z)\)

Questions for further discussion

• What are the functions in Example 2.1 geometrically?
• Does Equation 2.6 remain valid for all \(z\in\C\)? What is the correct modification?
• The two zeros in Figure 2.7 have a slightly different character. What is the difference between the zeros that might account for this?
• In (ex-Ch2_Fig_Image_Grid_1?) and (ex-Ch2_Ex1z?)} almost all the image gridlines meet at a right angle. The only exception is at \(f(0)\) in Example 2.6. Use polar coordinates to explain this behavior of \(f(z)=z^2\) at \(z=0.\)
• Is it always possible to extend the domain of definition of a complex function? Is this always sensible?
• Can you think of other branches of the logarithm?

2.1 Exercises

Exercise 2.1

Describe the image set of the complex function \(f(z)=\frac{1+z}{1-z}\) with domain \(D=\C\setminus\{1\}.\) In other words, determine the set of all \(w\in\C\) for which \(w=\frac{1+z}{1-z}\) has a solution \(z\in D.\)

Exercise 2.2

Sketch the following curves \(z(t)\) in the complex plane, where \(t\) is a real parameter.

1. \(z(t)= t(1+i)\) for \(0\leqslant t\leqslant 1\)
2. \(z(t)= \cos(t)+i\sin(t)\) for \(0\leqslant t\leqslant \pi\)
3. \(z(t)= \cos(t)-i\sin(t)\) for \(0\leqslant t \leqslant \pi/2\)
4. \(z(t)=\frac{1}{1+it}\) for \(t\in\R\)

Exercise 2.3

Let \(\log(z)\) be the principal branch of the logarithm. Compute \[\log(2i), \log(1+i), \log(-3i), \log(5).\]

Exercise 2.4

Describe the following complex functions geometrically. \[f(z)=3z, f(z)=iz, f(z)=\frac{(1+i)}{\sqrt{2}}z\]

Exercise 2.5

Determine a domain of definition for the following complex functions. \[f(z)=\frac{1}{z}, f(z)=\frac{1+z}{z-1}, f(z)=\frac{z^2-4}{z^2+2z}, f(z)=\frac{1}{\exp(z)}. \]

Exercise 2.6

Determine the domain of definition for \(f(z)=\frac{1}{\sin(z)}.\)

Exercise 2.7

Find all solutions to the following equations:

1. \(e^z=-1\),
2. \(\sin(z)=-i\)

Exercise 2.8

Prove that the composition \(f\circ f'\) of two Möbius transformations is again a Möbius transformation. If we associate to \(f, f'\) the matrices \[A=\begin{pmatrix} a&b\\ c&d \end{pmatrix}, A'=\begin{pmatrix} \alpha &\beta\\ \gamma& \delta \end{pmatrix} \in \M_{2\times 2}(\C),\]

show that \(f\circ f'\) is associated to the matrix product \(AA'.\)

Exercise 2.9

Draw the image grid for \(\exp\colon\C\to\C^\t.\)

Further resources

• https://youtu.be/NtoIXhUgqSk 5 ways to visualize a complex function
• Animation of Möbius transformations: https://youtu.be/0z1fIsUNhO4 and https://www-users.cse.umn.edu/~arnold/moebius/

Dowson, H. R. 1979. “Serge Lang, Complex Analysis (Addison-Wesley, 1977), Xi + 321 Pp., £1200.” Proceedings of the Edinburgh Mathematical Society 22 (1): 65–65. https://doi.org/10.1017/S0013091500027838.

Freitag, E., and R. Busam. 2009. Complex Analysis. Universitext. Springer Berlin Heidelberg. https://books.google.co.uk/books?id=3xBpS-ZKlgsC.

Howie, J. M. 2012. Complex Analysis. Springer Undergraduate Mathematics Series. Springer London. https://books.google.co.uk/books?id=0FZDBAAAQBAJ.

Jameson, G. J. O. 1987. “H. A. Priestley, Introduction to Complex Analysis (Oxford University Press, 1985), 197 Pp., £8.50.” Proceedings of the Edinburgh Mathematical Society 30 (2): 325–26. https://doi.org/10.1017/S0013091500028406.

Remmert, R. 1991. Theory of Complex Functions. Graduate Texts in Mathematics. Springer New York. https://books.google.co.uk/books?id=uP8SF4jf7GEC.

Thomson, B. S., J. B. Bruckner, and A. M. Bruckner. 2001. Elementary Real Analysis. Prentice-Hall. https://books.google.co.uk/books?id=6l_E9OTFaK0C.